Show that any continuous map $f: S^2 \rightarrow KB$ is null-homotopic.
Where $KB$ is the Klein bottle.
I am a bit confused on how to approach this, but I was thinking of using the universal cover of the klein bottle, $\mathbb{R}^2$ and covering space theory, because then every map $f: S^2 \rightarrow KB$ would have a lift that filters through the contractible spacee $\mathbb{R}^2$, so it seems like this should help. I'm not sure if this is correct, and if it is i'm having problems working out the details.
Best Answer
Take $\pi:\Bbb R^2\to KB$ the universal covering space. Because $S^2$ is simply connected you can lift $f$ to a map $\tilde{f}:S^2\to \Bbb R^2$. This one is null-homotopic, hence so is $\pi\circ \tilde{f}=f$.